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Q.
The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at $127^{\circ} C$ is
NTA AbhyasNTA Abhyas 2022
Solution:
Root mean square velocity
$v_{r m s}=\sqrt{\frac{3 R T}{M}}$
where $R$ is gas constant, $T$ the temperature and $M$ molecular weight.
Given, $M_{ N _{2}}=28, M_{ O _{2}}=32, T_{ O _{2}}=127 { }^{\circ} C =127+273=400\, K$
$\therefore \frac{v_{ O _{2}}}{v_{ N _{2}}}=\sqrt{\frac{T_{ O _{2}}}{M_{ O _{2}}} \times \frac{M_{ N _{2}}}{T_{ N _{2}}}}=\sqrt{\frac{400}{32} \times \frac{28}{T_{ N _{2}}}}=1 $
$\Rightarrow T_{ N _{2}}=350 K =77 { }^{\circ} C$